Applying the Bernoulli equation,

$p_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=p_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$

Here,  the height of the water in the tank is $h_1$,

$p_1$ is the pressure of water in the tank, and

$v_1$ is the velocity of the water. 

$h_2$ is the hole's height,

$p_2$ is the pressure at hole, and

$v_2$ is the water's pace at the hole and 

$\rho$ represents the water density.

There acts the same atmospheric pressure at the top of the tank and at the hole, thus $p_1=p_2$.

We will neglect the water velocity at the top since the tank is large as its velocity is much smaller than the velocity at the hole.

Now, the Bernoulli equation becomes, $\rho g{h}_1=\frac{1}{2}\rho v_2^2+\rho g{h}_2$

$\Rightarrow v_2=\sqrt{2g\left(h_1-h_2\right)}$

$\Rightarrow v_2=\sqrt{2\times 9.81\times \left(0.2\right)}=1.98 \mathrm{m} {\mathrm{s}}^{-1}$

The drainage rate is $A_2{v}_2=6\times {10}^{-4}\times 1.98$

$\Rightarrow A_2{v}_2=1.18\times {10}^{-3}=1.2\times {10}^{-3} {\mathrm{m}}^3 {\mathrm{s}}^{-1}$